Description
You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.
Input
The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.
Output
You need to answer all Q commands in order. One answer in a line.
Sample Input
10 5 1 2 3 4 5 6 7 8 9 10 Q 4 4 Q 1 10 Q 2 4 C 3 6 3 Q 2 4
Sample Output
4 55 9 15
Hint
The sums may exceed the range of 32-bit integers.
题目大意：给出一些数和一些操作，q表示询问，c表示修改。
思路：纯线段树
1 /*
2 * Author: Joshua
3 * Created Time: 2014年07月16日 星期三 15时19分46秒
4 * File Name: poj3468.cpp
5 */
6 #include<cstdio>
7 #include<algorithm>
8 #define maxn 100005
9 #define L(x) (x<<1)
10 #define R(x) (x<<1 |1)
11 typedef long long LL;
12 using namespace std;
13 struct node
14 {
15 int l,r;
16 LL sum,mark;
17 } e[maxn <<2];
18 int n,m;
19 int a[maxn];
20 LL ans;
21 void built(int t,int l,int r)
22 {
23 e[t].l=l;
24 e[t].r=r;
25 e[t].mark=0;
26 if (l==r)
27 {
28 e[t].sum=a[e[t].l];
29 return;
30 }
31 int mid=l+r >>1;
32 built(L(t),l,mid);
33 built(R(t),mid+1,r);
34 e[t].sum=e[L(t)].sum+e[R(t)].sum;
35 }
36
37 void add(int t,int l,int r,int v)
38 {
39 if (l==e[t].l && r==e[t].r)
40 {
41 e[t].mark+=v;
42 return;
43 }
44 e[t].sum+=(r-l+1)*v;
45 int mid=e[t].l+e[t].r >>1;
46 if (r>mid) add(R(t),max(mid+1,l),r,v);
47 if (l<=mid) add(L(t),l,min(mid,r),v);
48 }
49
50 void getSum(int t,int l,int r)
51 {
52 ans+=e[t].mark*(r-l+1);
53 if (e[t].l==l && e[t].r==r)
54 {
55 ans+=e[t].sum;
56 return;
57 }
58 int mid=e[t].l+e[t].r >>1;
59 if (r>mid) getSum(R(t),max(mid+1,l),r);
60 if (l<=mid) getSum(L(t),l,min(mid,r));
61 }
62 void solve()
63 {
64 for (int i=1;i<=n;++i)
65 scanf("%d",&a[i]);
66 built(1,1,n);
67 char c;
68 int x,y,v;
69 for (int i=1;i<=m;++i)
70 {
71 scanf("%c",&c);
72 scanf("%c%d%d",&c,&x,&y);
73 if (c==‘Q‘)
74 {
75 ans=0;
76 getSum(1,x,y);
77 printf("%lld\n",ans);
78 }
79 else
80 {
81 scanf("%d",&v);
82 add(1,x,y,v);
83 }
84 }
85 }
86 int main()
87 {
88 while (scanf("%d%d",&n,&m)==2)
89 solve();
90
91 return 0;
92 }
poj3468 A Simple Problem with Integers,布布扣,bubuko.com